Reciprocal Space
Removed due to copyright restrictions.
Please see:
Massa, Werner. Crystal Structure Determination. 2nd ed. Translated into English by R. O. Gould.
New York, NY: Springer, 2004. ISBN: 3540206442.
The reflection h, k, l is generated by diffraction of the X-ray beam at the Bragg
plane set h, k, l, which intersects the three edges of the unit cell at 1/h, 1/k and 1/l.
Sets of planes in real space (with spacing d) correspond to points in reciprocal
space (distance d* from the origin). The vector d* is perpendicular to the Bragg
planes and has the length |d*| = 2sinθ/λ.
A reflection is visible when the corresponding set of Bragg planes is in reflex
position, that is when Bragg’s law is fulfilled. In an alternative description:
a reflection is visible when the corresponding scattering vector s = d* intersects
with the Ewald sphere.
Reciprocal Space
Removed due to copyright restrictions.
Please see:
Massa, Werner. Crystal Structure Determination. 2nd ed. Translated into English by R. O. Gould.
New York, NY: Springer, 2004. ISBN: 3540206442.
The reflections form a lattice in reciprocal space. Reciprocal unit cell: a*, b*, c*
The dimensions and angles of the reciprocal cell are inversely proportional to the
real space cell: if the unit cell doubles, the space between the X-ray reflections
will be reduced by factor two.
Reciprocal Space
001
011
c*
111
101
For orthorhombic tetragonal and cubic unit cells:
a* = 1/a
b* = 1/b
c* = 1/c
α* = β* = γ* = α = β = γ = 90°
Triclinic more complicated:
β*
α*
000
γ*
b* 010
a*
100
110
Courtesy of George M. Sheldrick. Used with permission.
1/V = V* = a*b*c* [ 1 – cos2α* – cos2β* – cos2γ* + 2cosα*cosβ*cosγ* ]½
a = b*c*sinα* / V* and cosα = ( cosβ*cosγ* – cosα* ) / ( sinβ*sinγ* )
Same thing for b, c, cosβ, cosγ and for a*, cosα* etc.
The Reciprocal Lattice: Ewald Construction
hkl reciprocal
lattice point
Diffracted beam
Detector
s
Q
θ
Incident
beam
θ
C
hkl reflection
P
θ
2θ
s
d
O
Crystal
hkl lattice
planes
Ewald sphere
with radius r =
1/λ
Reciprocal lattice
Intensities of the Reflections
With the help of Braggs law and the Ewald construction, we can
calculate the place of a reflection on the detector, provided we know
the unit cell dimensions. Indeed, the position of a spot is determined
alone by the metric symmetry of the unit cell. The intensity of a spot,
however, depends on the contents of the unit cell (and, of course, on
exposure time, crystal size, etc.).
The intensity of a reflection depends, among other things, on the
population of the corresponding set of Bragg planes with atoms. If
there are many atoms on a plane, the corresponding reflection is
strong, if the plane is empty, the reflection is weak or absent.
Other factors are the thermal motion of the atoms and the atomic
radius.
Atomic Form Factors
X-rays are scattered by the electrons of an atom, hence the point
atom approximation does not hold too well. Ergo: Bragg planes are
not mathematical planes but have a fuzzy thickness, which means
there is a path difference occurring within each individual Bragg plane.
Displacement of the place of diffraction from the ideal position of δ
leads to a path difference of δ·2π/d. This has a weakening effect on
the intensity of the corresponding reflection. This effect is a function of
the distance d between the Bragg planes: it becomes stronger with
smaller distances (i.e. at higher resolution).
δ
d
½∆
½∆
Atomic Form Factors
In a first approximation the atomic scattering factor is dependent on
the atom number (number of electrons). With increasing resolution
(that is decreasing Bragg plane distance, equivalent to higher
scattering angles 2Θ) the scattering factor becomes smaller.
Removed due to copyright restrictions.
Please see:
Massa, Werner. Crystal Structure Determination. 2nd ed. Translated into English by R. O. Gould.
New York, NY: Springer, 2004, pp. 34. ISBN: 3540206442.
Fig. 5.2
Atomic Displacement Factors
Atoms not only have a radius, they also move. With the displacement
of an atom from its ideal position on the Bragg plane, an additional
path difference is added.
The same principle as with the atomic form factors applies (stronger
weakening at high scattering angles), but this effect is stronger when
the thermal motion is higher.
Removed due to copyright restrictions.
Please see:
Massa, Werner. Crystal Structure Determination. 2nd ed. Translated into English by R. O. Gould.
New York, NY: Springer, 2004, p. 35. ISBN: 3540206442.
Fig. 5.3
 − 2π 2 u 2 
f ' = f ⋅ exp 

2

 d
Atomic Displacement Factors
Atoms not only have a radius, they also move. With the displacement
of an atom from its ideal position on the Bragg plane, an additional
path difference is added.
The same principle as with the atomic form factors applies (stronger
weakening at high scattering angles), but this effect is stronger when
the thermal motion is higher.
 − 2π 2 u 2 
f ' = f ⋅ exp 

2
d


f’: atomic scattering factor; f: atomic form factor;
u: mean vibration amplitude; d: Bragg plane spacing
U = u2 is the atomic displacement factor or temperature factor.
Atomic Displacement Factors
 − 2π 2 u 2 
f ' = f ⋅ exp 

2

 d
We know: d = 1/d*= 1/(2sinθ/λ)
and: U = u2. Thus:
[
f ' = f ⋅ exp − 2π 2Ud *2
]
2

sin
θ
2
f ' = f ⋅ exp − 8π U 2 
λ 

Substitue: B = 8π2U:

sin 2 θ 
f ' = f ⋅ exp − B 2 
λ 

f’: atomic scattering factor; f: atomic form factor; u: mean vibration amplitude;
U: temperature factor; d: Bragg plane spacing, d* scattering vector; B: Debye-Weller Factor.
Structure Factors
So far we were talking about single atoms at 0,0,0 in the unit cell.
Lets assume two identical atoms: atom 1 is at 0,0,0 and atom 2 is at x2, y2, z2.
Contribution of the blue atoms to the 1 1 0 reflection is same as that of
the red atoms, however with a phase shift.
110
planes
b
y2
x2
0 ≤ ∆Φ ≤ 2π
a
Structure Factors
Φ i = ∆Φ i ( a ) + ∆Φ i ( b ) + ∆Φ i ( c ) = 2π (hxi + kyi + lzi )
This makes the structure factor a complex number:
Fi = f i ⋅ exp[iΦ i ] = f i (cos Φ i + i sin Φ i )
Not the same i !
Structure Factors
Φ i = ∆Φ i ( a ) + ∆Φ i ( b ) + ∆Φ i ( c ) = 2π (hxi + kyi + lzi )
This makes the structure factor a complex number:
Fi = f i ⋅ exp[iΦ i ] = f i (cos Φ i + i sin Φ i )
Every atom i in the unit cell contributes to every structure factor F(hkl)
(that is reflection) according to its position in the cell and its chemical
nature (different values for fi !):
F ( hkl ) = ∑ f i [cos 2π (hxi + kyi + lzi ) + i sin 2π (hxi + kyi + lzi )]
i
Structure Factors
F ( hkl ) = ∑ f i [cos 2π (hxi + kyi + lzi ) + i sin 2π (hxi + kyi + lzi )]
i
Argand plane
i
Φ5
Φ4
F4
F3 Φ3
F2
F
Φ2
Φ1
F1
r
Structure Factors
F ( hkl ) = ∑ f i [cos 2π (hxi + kyi + lzi ) + i sin 2π (hxi + kyi + lzi )]
i
FT
ρ x , y , z = 1V ∑ Fhkl ⋅ exp[− i 2π (hx + ky + lz )]
hkl
The electron density at every given place in the unit cell (real space) can be
calculated from the equation above.
Note that xi, yi, zi in the structure factor equation refer to atomic coordinates,
while x, y, z in the electron density equation refer to arbitrary places
anywhere in the unit cell. Of course, the electron density at the place of an
atom is supposed to be much higher than the electron density at a place
where there is “nothing”.
Electron Density
ρ x , y , z = 1V ∑ Fhkl ⋅ exp[− i 2π (hx + ky + lz )]
hkl
1.7 Å
It is sufficient to calculate the electron density at
a number of grid points within the unit cell and
extrapolate between the points.
0.9 Å
Courtesy of George M. Sheldrick. Used with permission.
Resolution
The maximum resolution d can be calculated from the Bragg equation:
d = λ / 2 sin(θmax)
This d corresponds approximately to the distance
between two atoms (or more generally points in
space), which can still be resolved from one
another in an electron density map calculated from
data at that particular resolution.
For centrosymmetric structures one would expect
ca. 80/d3 independent reflections per non-hydrogen
atom. For non-centrosymmetric structures only half
that number (assuming Friedel’s law is true). That
means, for a protein structure at 1.5 Å resolution
we have ca. 12 reflections per atom. At 2.5 Å only
ca. 2.6 reflections per atom. There are many
protein structures at 3 Å (1.5 spots per atom) and
worse…
~d
Courtesy of George M. Sheldrick.
Used with permission.
Resolution
FT
FT of a duck
FT
FT of a duck
with high
resolution data
removed
FT
FT
A duck
A low
resolution
duck
Courtesy of Kevin Cowtan. http://www.ysbl.york.ac.uk/~cowtan/ Used with permission.
Resolution
0.9 Å
2.5 Å
1.5 Å
4.0 Å
Electron density contour plots of the same molecule at different resolutions.
Courtesy of George M. Sheldrick. Used with permission.
Calculating Electron Density
ρ x , y , z = 1V ∑ Fhkl ⋅ exp[− i 2π (hx + ky + lz )]
hkl
To make this equation work we need the volume of the unit cell (easy), a
dataset with the intensities for h, k and l (also easy) and the phase of every
structure factor (not quite so easy).
We can measure the intensities and determine the unit cell from the
locations of the reflections, but it is very difficult to actually measure the
relative phase of the reflections. This is known as the “crystallographic
phase problem”.
MIT OpenCourseWare
http://ocw.mit.edu
5.069 Crystal Structure Analysis
Spring 2008
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